Template specialization based on inherit class

Question

I want to make this specialized w/o changing main. Is it possible to specialize something based on its base class? I hope so.

-edit-

I'll have several classes that inherit from SomeTag. I don't want to write the same specialization for each of them.

class SomeTag {};
class InheritSomeTag : public SomeTag {};

template <class T, class Tag=T>
struct MyClass
{
};

template <class T>
struct MyClass<T, SomeTag>
{
    typedef int isSpecialized;
};

int main()
{
    MyClass<SomeTag>::isSpecialized test1; //ok
    MyClass<InheritSomeTag>::isSpecialized test2; //how do i make this specialized w/o changing main()
    return 0;
}

Answer 1

This article describes a neat trick: http://www.gotw.ca/publications/mxc++-item-4.htm

Here's the basic idea. You first need an IsDerivedFrom class (this provides runtime and compile-time checking):

template<typename D, typename B>
class IsDerivedFrom
{
  class No { };
  class Yes { No no[3]; }; 

  static Yes Test( B* ); // not defined
  static No Test( ... ); // not defined 

  static void Constraints(D* p) { B* pb = p; pb = p; } 

public:
  enum { Is = sizeof(Test(static_cast<D*>(0))) == sizeof(Yes) }; 

  IsDerivedFrom() { void(*p)(D*) = Constraints; }
};

Then your MyClass needs an implementation that's potentially specialized:

template<typename T, int>
class MyClassImpl
{
  // general case: T is not derived from SomeTag
}; 

template<typename T>
class MyClassImpl<T, 1>
{
  // T is derived from SomeTag
  public:
     typedef int isSpecialized;
}; 

and MyClass actually looks like:

template<typename T>
class MyClass: public MyClassImpl<T, IsDerivedFrom<T, SomeTag>::Is>
{
};

Then your main will be fine the way it is:

int main()
{
    MyClass<SomeTag>::isSpecialized test1; //ok
    MyClass<InheritSomeTag>::isSpecialized test2; //ok also
    return 0;
}

Answer 2

Well, the article in the answer above appeared in February 2002. While it works, today we know there are better ways. Alternatively, you can use enable_if :

template<bool C, typename T = void>
struct enable_if {
  typedef T type;
};

template<typename T>
struct enable_if<false, T> { };

template<typename, typename>
struct is_same {
    static bool const value = false;
};

template<typename A>
struct is_same<A, A> {
    static bool const value = true;
};

template<typename B, typename D>                                 
struct is_base_of {                                                       
    static D * create_d();                     
    static char (& chk(B *))[1]; 
    static char (& chk(...))[2];           
    static bool const value = sizeof chk(create_d()) == 1 &&  
                              !is_same<B    volatile const, 
                                       void volatile const>::value;
};

struct SomeTag { };
struct InheritSomeTag : SomeTag { };

template<typename T, typename = void>
struct MyClass { /* T not derived from SomeTag */ };

template<typename T>
struct MyClass<T, typename enable_if<is_base_of<SomeTag, T>::value>::type> {
    typedef int isSpecialized;
};

int main() {
    MyClass<SomeTag>::isSpecialized test1;        /* ok */
    MyClass<InheritSomeTag>::isSpecialized test2; /* ok */
}

Answer 3

And the short version now, 2014, using C++-11:

#include <type_traits>

struct SomeTag { };
struct InheritSomeTag : SomeTag { };

template<typename T, bool = std::is_base_of<SomeTag, T>::value>
struct MyClass { };

template<typename T>
struct MyClass<T, true> {
    typedef int isSpecialized;
};

int main() {
    MyClass<SomeTag>::isSpecialized test1;        /* ok */
    MyClass<InheritSomeTag>::isSpecialized test2; /* ok */
}

Answer 4

In your case, the only way that I see would be to explicitly specialize MyClass for InheritSomeTag . However, the SeqAn paper proposes a mechanism called “template sublassing” that does what you want – albeit with a different inheritance syntax, so the code isn't compatible with your current main function.

// Base class
template <typename TSpec = void>
class SomeTag { };

// Type tag, NOT part of the inheritance chain
template <typename TSpec = void>
struct InheritSomeTag { };

// Derived class, uses type tag
template <typename TSpec>
class SomeTag<InheritSomeTag<TSpec> > : public SomeTag<void> { };

template <class T, class Tag=T>
struct MyClass { };

template <class T, typename TSpec>
struct MyClass<T, SomeTag<TSpec> >
{
    typedef int isSpecialized;
};

int main()
{
    MyClass<SomeTag<> >::isSpecialized test1; //ok
    MyClass<SomeTag<InheritSomeTag<> > >::isSpecialized test2; //ok
}

This certainly looks strange and is very cumbersome but it allows a true inheritance mechanism with polymorphic functions that is executed at compile time. If you want to see this in action, have a look at some SeqAn examples .

That being said, I believe that SeqAn is a special case and not many applications would profit from this extremely difficult syntax (deciphering SeqAn-related compiler errors is a real pain in the *ss!)

Answer 5

Using concepts and the requires keyword from C++20 is an even simpler and more expressive way to do this without having to introduce a redundant boolean non-type template parameter like in C++11:

// C++20:
#include <concepts>
#include <iostream>

struct SomeTag { };
struct InheritSomeTag : SomeTag { };

template<typename T>
struct MyClass 
{ 
    void Print()
    {
        std::cout << "Not derived from someTag\n";
    }
};

// std::derived_from is a predefined concept already included in the STL
template<typename T>
    requires std::derived_from<T, SomeTag> 
struct MyClass<T> 
{
    void Print()
    {
        std::cout << "derived from someTag\n";
    }
};

int main() 
{
    MyClass<InheritSomeTag> test1;
    test1.Print();      // derived from someTag
    MyClass<int> test2;        
    test2.Print();      // Not derived from someTag

    // Note how even the base tag itself returns true from std::derived_from:
    MyClass<SomeTag> test3;          
    test3.Print();      // derived from someTag 

}